Problem: $g(x)=\dfrac{4}{1-2x}$ We know that $g(x)=4+8x+16x^2+32x^3+...$ for $x\in\left(-\dfrac12,\dfrac12\right)$. Using this fact, find the power series for $f(x)=\dfrac{8}{(1-2x)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $ 16+64x+192x^2+...$ (Choice B) B $ -16-64x-192x^2+...$ (Choice C) C $-2-4x-8x^2+...$ (Choice D) D $ 32+128x+384x^2+...$ (Choice E) E $ 8+32x+96x^2+...$
Explanation: First, notice that the derivative of $ ~~g(x)=\frac{4}{1-2x}~~$ is $~~\frac{8}{{{(1-2x)}^{2}}}\,$. This is precisely the function $~f(x)~$ that we are looking for an approximation to. $ g(x)=\frac{4}{1-2x}=4\cdot \left(\frac{1}{1-2x}\right)$ From our knowledge of geometric series, we know the following way to expand $~~\dfrac{1}{1-2x}\,$. $\frac{1}{1-2x}=1+2x+4x^2+8x^3+16x^4+...$ Hence, $ g(x)=4\cdot\left(\frac{1}{1-2x}\right)=4+8x+16x^2+32x^3+64x^4+...$ Taking the derivative of both sides leads to our result. $ g\,^\prime(x)=f(x)=\frac{8}{(1-2x)^2}=8+32x+96x^2+256x^3+...$